A charge q is enclosed by a Gaussian spherical surface of radius R. If the radius is reduced to half of R, the outward electric flux will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → the same as its initial value
According to Gauss's law:
$\Phi_E = \frac{q_\text{enclosed}}{\epsilon_0}$
The electric flux depends only on the total charge enclosed and not on the radius of the Gaussian surface.
∴ Reducing the radius to half, the outward electric flux remains the same